1/5/2024 0 Comments Flat space curved space![]() So any spacetime that does not have a chart in which the metric coefficients are constant must be curved. So any spacetime with vanishing Riemann tensor must have the same geometry as Minkowski spacetime, and must therefore admit a coordinate chart with constant metric coefficients. But this is not possible the easiest way to see this is to observe that all the invariants that determine the spacetime geometry are functions of the Riemann tensor, so if the Riemann tensor vanishes identically, all the invariants must also vanish. In flat coordinates, the orbiting body is attracted to where the opposing body is at that moment with the force carried instantaneously, whereas in curved coordinates, the orbiting body is attracted to where the opposing body was a moment earlier with the force carried at the speed of light. In other words, there would have to be some other spacetime that does not admit a coordinate chart in which the metric coefficients are constant, but still has a vanishing Riemann tensor (which is what "flat spacetime" means). Knowing that, we can easily see that your proposition is true as well, as follows: for the converse to be true, as above, but your proposition to be false, there would have to be at least two flat spacetimes, because we know Minkowski spacetime does have a coordinate chart with constant metric coefficients, and is therefore flat by the converse proposition above (as of course we know it is). The converse is certainly true: if there does exist a coordinate chart in which the metric coefficients are constant, then spacetime is flat (since obviously all derivatives of the metric coefficients are zero, and therefore the Riemann tensor vanishes identically). So shouldnt the statement instead be that if there does not exist coordinate system in which the components of the metric are not functions of space-time, the space-time is curved from which it follows that proper time is location dependent.Īm I correct when I say that if there does not exist coordinate system in which the components of the metric are not functions of space-time, the space-time is curved? So from the above statement I would conclude that proper time also varies from location to location in SR. In flat space am I correct in thinking that the components also can depend upon space-time, e.g use of polar coordinates, and the only case where it doesnt is cartesian coordinates? I believe it is always the case that the comtponents of a metric in GR are functions of space-time, and so I understand that proper time will vary for observers dependent upon location, but, It then makes the comment that proper time depends on the the observers location (compared to SR where it doesn't). ![]() Put another way, If 1, the universe is flat. It uses a (+,-,-,-) signature and so #proper time=ds^#. The density parameter is the average density of the universe divided by the critical energy density, that is, the mass energy needed for a universe to be flat. I'm reading An Intro to GR, Hughston and Tod, it says that in GR the idea is that the geometry of st varies from point to point, represented by allowing the metric to vary over space-time.
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